Question

The line $y=x+1$ is a tangent to the curve $y^2=4x$ at the point.

• A. $(1,-2)$
• B. $(2,1)$
• C. $(-1,2)$
• D. $(1,2)$

Answer 1

The correct option is D $(1,2)$

Let the point where tangent is drawn be $(h,k)$
Given curve is $y^2=4x$
Diffrentiating w.r.t. $x$,
$\frac{d\left(y^2\right)}{dx}=\frac{d\left(4x\right)}{dx}$
$\Rightarrow 2y\times \frac{dy}{dx}=4$
$\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
Slope of tangent at $(h,k)$ is
$\Rightarrow \frac{dy}{dx}=\frac{2}{k}$
Given tangent is $y=x+1$
Slop of this line is $1$, so $\frac{dy}{dx}=1$
$\Rightarrow \frac{2}{k}=1$
$\Rightarrow k=2$
And $y^2=4x$
$\Rightarrow (k{)}^2=4h$
$\Rightarrow 2^2=4h$
$\Rightarrow h=1$
So, the required point is $(1,2)$
Hence,correct answer is $\left(A\right)$.